When we talked about resistors in series, we saw that the total resistance of the circuit was equal to the sum of the resistances of the individual components (click >> here << to quickly recap if you want to).


In parallel circuits things are not quite so straightforward, if you have two resistors in parallel then their total resistance is less than the resistance of the smaller of the two.





The values in the diagram above have been taken from a circuit simulator showing three resistors in parallel, attached to a 12 V potential difference drawing 7.9 A in total through the circuit.


Resistor number one (8 ohms) draws 1.5 A from the total which is easily calculated since if V=IR then I = V/R = 12/8 = 1.5.


By the same reasoning, Resistor number two draws 2.4 A and Resistor number three draws 4 A. Since the meters attached to the circuit show an overall current of 7.9 A from a potential difference of 12 V we can conclude that the overall resistance of the circuit is in fact R = V/I = 12/7.9 or approximately 1.52 ohms which is indeed smaller than the smallest resistor in the circuit (3 ohms). 


Therefore the line which states that two or more resistors in parallel will give a total resistance less than the resistance of the smallest is shown to be correct.




This is the same circuit as above but shown in the simulator. If we add a fourth resistor to the setup the overall resistance will be even lower, because adding another loop gives the current another path to flow along, and there will be an increase in current on the circuit overall:



In this case, adding a resistor with a resistance of 2 ohms draws another 6 A from the circuit, the drain on the circuit increases to 13.9 A with a new overall resistance of 0.86 ohms (again, lower than the smallest value of 2 ohms).